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TEKS: G5B, G5D, G11A, G11B, G11C
The student will use numeric patterns to make
generalizations about geometric properties including
properties of polygons and ratios in similar figures.
The student will identify and apply patterns from
right triangles.
The student will use and extend similarity properties
and transformations to explore and justify
conjectures about geometric figures.
The student will use ratios to solve problems
involving similar figures.
The student will develop, apply, and justify right
triangle ratios.
Remember
In a right triangle, an altitude drawn from the
vertex of the right angle to the hypotenuse forms
two right triangles.
Example: 1
Write a similarity statement
comparing the three triangles.
Sketch the three right triangles with the angles of the
triangles in corresponding positions.
W
Z
By Theorem 8-1-1, ∆UVW ~ ∆UWZ ~ ∆WVZ.
Example: 2
Write a similarity statement
comparing the three triangles.
Sketch the three right triangles with the angles of
the triangles in corresponding positions.
By Theorem 8-1-1, ∆LJK ~ ∆JMK ~ ∆LMJ.
extremes
means
Consider the proportion
means
extremes
In this case, the means of the proportion
are the same number, and that number is
the geometric mean of the extremes.
The geometric mean of two positive
numbers is the positive square root of
their product.
So the geometric mean of a and b is the
positive number x such
that
,or x2 = ab.
Example: 3
Find the geometric mean of each pair of numbers. If
necessary, give the answer rounded to the nearest tenth.
4 and 25
Let x be the geometric mean.
x2 = (4)(25) = 100
x = 10
Def. of geometric mean
Find the positive square root.
Example: 4
Find the geometric mean of each pair of numbers. If
necessary, give the answer rounded to the nearest tenth.
5 and 30
Let x be the geometric mean.
x2 = (5)(30) = 150
Def. of geometric mean
x = 150
Find the positive square root.
x≈12.2
Helpful Hint
Once you’ve found the unknown side lengths, you
can use the Pythagorean Theorem to check your
answers.
Set Up # 1
B
altitude
A
segment1D
C
segment2
segment1
altitude

altitude segment2
Set Up # 2
B
leg
A
C
adj. D
hypotenuse
adjacent segment
leg

leg
hypotenuse
Set Up # 3
B
leg
A
D
adj.
C
hypotenuse
adjacent segment
leg

leg
hypotenuse
Example: 5
B
Find BD
2
BD

BD
8
(BD)2  16
BD  4
A
2
D
8
C
Example: 6
B
Find BD
3
BD

BD
8
(BD)2  24
BD  24
BD  4 6
BD  2 6
BD  4.9
A 3 D
8
C
Example: 7
B
Find DC and AC
10
5
10

10 DC
5(DC )2  100
DC  20
AC  20  5  25
A 5 D
C
Example: 8
Find AD and AC
AD
3 2

12
3 2
12( AD) 9 4

12
12
AD  1.5
AC  12  1.5  13.5
B
3 2
A
D
12
C
Example: 9
B
Find BC
5
BC

BC
9
(BC )2  45
BC  9 5  3 5
C
5
D4 A
Example: 10
B
Find CD
9
12

12 9  DA
81 9DA  144
DA  7
12
C
D9
A
Example: 11
B
Find BC
3 7

7
3 7
3 7
CD
D 7 A
C
9 49  7(CD)
63  7(CD)
CD  9
9
BC

BC 16
(BC )2  144
BC  12
Example: 12
B
Find AD
12
3 2
a2  b2  c 2
2
2
2
(3 2)  AD  12
C
18  AD2  144
AD2  126
AD  126  9 14  3 14
AD  11.22
D
A
Example: 13
To estimate the height of a
Douglas fir, Jan positions herself
so that her lines of sight to the top
and bottom of the tree form a 90º
angle. Her eyes are about 1.6 m
above the ground, and she is
standing 7.8 m from the tree.
What is the height of the tree to
the nearest meter?
1.6 7.8

7.8
x
x  38.025
The tree is about 38025+1.6 = 39.625,
or 40m tall
Example: 14
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A surveyor positions
himself so that his line of
sight to the top of a cliff
and his line of sight to the
bottom form a right angle
as shown.
What is the height of the
cliff to the nearest foot?
5.5 28

28
x
x  142.5
The cliff is about 142.5 + 5.5, or 148ft high
Example: 15 Pre AP
Find CA and CD
B
12
3 2 12

12
CA
3 2(CA) 144

3
3
CA 
48
2

48 2
CA 
2
C
D3
2
2
CA  24 2
CD  24 2  3 2  21 2
2
A